In a set of five consecutive integers, the largest integer is less than twice the average of the five integers. What is the smallest integer that could be in the set?
Answer: Let the integers be $n$, $n+1$, $n+2$, $n+3$ and $n+4$.  Their average is $n+2$, so we have $$n+4<2(n+2) \Rightarrow n+4<2n+4 \Rightarrow 0<n.$$Thus $n$ is at least $\boxed{1}$.